159. Find Minimum in Rotated Sorted Array 【medium】

159. Find Minimum in Rotated Sorted Array 【medium】

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

 Notice

You may assume no duplicate exists in the array.

Have you met this question in a real interview? 

Yes

Example

Given [4, 5, 6, 7, 0, 1, 2] return 0

解法一：

class Solution {
public:
    /*
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    int findMin(vector<int> nums) {
        int start = 0;
        int end = nums.size() - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            
            //1 2 3 4 5 6
            if (nums[mid] > nums[start] && nums[mid] < nums[end]) {
                end = mid;
            }
            // 3 4 5 6 1 2
            else if (nums[mid] > nums[start] && nums[mid] > nums[end]) {
                start = mid;
            }
            // 5 6 1 2 3 4
            else if (nums[mid] < nums[start] && nums[mid] < nums[end]) {
                //start = mid;
                end = mid;
            }
        }
        
        return nums[start] > nums[end] ? nums[end] : nums[start];
    }
};

注意分数组的奇偶去考查。比如[1, 2, 3, 4, 5]和[1, 2, 3, 4, 5, 6]是不同的。

解法二：

public class Solution {
    /**
     * @param nums: a rotated sorted array
     * @return: the minimum number in the array
     */
    public int findMin(int[] nums) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int start = 0, end = nums.length - 1;
        int target = nums[nums.length - 1];
        
        // find the first element <= target
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] <= target) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if (nums[start] <= target) {
            return nums[start];
        } else {
            return nums[end];
        }
    }
}

大神解法就是给力！